Consider a spherical soap bubble with an initial radius of 30 cm. A small thread making a circular loop of radius 1 mm is placed on the surface. Now the soap film inside the loop is pierced. As a result the air inside the soap bubble starts flowing out. Assume that the air flowing out is an ideal incompressible fluid that follows Bernoulli's theorem. Air temperature inside and outside the bubble t = 27°C. Surface tension of the soap solution at this temperature S = 0.005 N/m. Average molecular mass of air taken as 30 gm/mol. Atmospheric pressure = 105 Pa.

Engineering

Physics

Surface Tension

Question

Consider a spherical soap bubble with an initial radius of 30 cm. A small thread making a circular loop of radius 1 mm is placed on the surface. Now the soap film inside the loop is pierced. As a result the air inside the soap bubble starts flowing out. Assume that the air flowing out is an ideal incompressible fluid that follows Bernoulli's theorem. Air temperature inside and outside the bubble t = 27°C. Surface tension of the soap solution at this temperature S = 0.005 N/m. Average molecular mass of air taken as 30 gm/mol. Atmospheric pressure = 10⁵ Pa.

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Linked Question 1

What is the initial velocity of the air just outside the hole if the velocity of flow inside the bubble can be neglected?

$\frac{2}{3} m / s$

$\frac{1}{3} m / s$

$\frac{1}{2} m / s$

$\frac{1}{3 \sqrt{2}} m / s$

$\frac{ρ_{0} + \frac{4 s}{r}}{ρg} + 0 = \frac{ρ_{0}}{ρg} + \frac{v^{2}}{2 g}$

$v^{2} = \frac{8 s}{ρr}$

$ρ = \frac{PM}{RT} = \frac{10^{5} \times 30 \times 10^{- 3}}{\frac{25}{3} \times 300} = \frac{6}{5} kg / m^{3}$

$y = \sqrt{\frac{8 \times 0 .005}{\frac{6}{5} \times 0 .3}} = \sqrt{\frac{\frac{2}{5}}{\frac{18}{5}}} = \frac{1}{3} m / s$

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Linked Question 2

Estimate the time at which all the air will come out of the bubble?

$\frac{341}{2} \times 10^{4} \sec$

$\frac{648}{7} \times 10^{3} \sec$

$\frac{248}{3} \times 10^{3} \sec$

144 × 10³ sec.

$Av = \frac{dv}{dt}$

${πR}^{2} \times \sqrt{\frac{8 s}{ρr}} = \frac{d}{dt} (\frac{4}{3} {πr}^{3}) = 4 π^{2} \frac{dr}{dt}$

$\frac{R^{2}}{4} \sqrt{\frac{8 s}{P}} \int r^{5 / 2} dr$

$\frac{R^{2}}{4} \times \sqrt{\frac{8 s}{ρ}} \times t = {\frac{2 r^{7 / 2}}{7}|}_{0 .3}^{0}$

$t = \frac{4}{R^{2}} \sqrt{\frac{ρ}{8 s}} \times \frac{2}{7} \times r^{7 / 2}$

$= \frac{8}{7 \times 10^{- 6}} \sqrt{\frac{6}{40 \times 5 \times 10^{- 3}}} \times {(\frac{3}{10})}^{7 / 2}$

$= \frac{8}{7} \times 10^{6} \sqrt{30} \times \sqrt{\frac{3}{10}} \times \frac{27}{1000} = \frac{648}{7} \times 10^{3} \sec$

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